Question

Find the area bounded by the curve y2 = 12x and the line x = 3, in the first
quadrant.

#Class 12th

Don't spam ⭐​

Answer 1

EXPLANATION.

Area bounded by the curve : y² = 12x and the line x = 3. in the first quadrant.

As we know that,

Curve : y² = 12x.

It is a curve of a parabola.

General equation of parabola : y² = 4ax.

⇒ y = √12x.

$\implies \displaystyle \int\limits^3_0 {y} \, dx$

$\implies \displaystyle \int\limits^3_0 {\sqrt{12x} } \, dx$

$\implies \displaystyle \sqrt{12} \int\limits^3_0 {\sqrt{x} } \, dx$

$\implies \displaystyle \sqrt{12} \int\limits^3_0 (x)^{1/2} dx$

$\implies \displaystyle \sqrt{12} \bigg[ \dfrac{(x)^{1/2 + 1} }{1/2 + 1} \bigg]_{0}^{3}$

$\implies \displaystyle \sqrt{12} \bigg[\dfrac{(x)^{3/2} }{3/2} \bigg]_{0}^{3}$

$\implies \displaystyle \sqrt{12} \times \dfrac{2}{3} \bigg[ (x)^{3/2} \bigg]_{0}^{3}$

$\implies \displaystyle \dfrac{2\sqrt{12} }{3} \bigg[ (3)^{3/2} \ - (0)^{3/2} \bigg]$

$\implies \displaystyle \dfrac{2\sqrt{12} }{3} \bigg[ (3)^{3/2} \bigg]$

$\implies \displaystyle \dfrac{2\sqrt{12} }{3} \times 3\sqrt{3}$

$\implies \displaystyle 2\sqrt{12} \times \sqrt{3} = 4\sqrt{3} \times \sqrt{3} = 12$